Given set data (from the passage): \( |M|=55,\ |L|=38,\ |D|=22,\ |E|=50 \);
\(|M\cap E|=30,\ |L\cap E|=15,\ |M\cap L|=20,\ |M\cap L\cap E|=5\);
\(|D\cap M|=5,\ |D\cap L|=5,\ |D\cap M\cap L|=5\); and \(E\cap D=\varnothing\).
[2mm]
Inside \(E\) (no \(D\)):
Only \(E=50-30-15+5=10\),\quad \(ME\text{-only}=30-5=25\),\quad \(LE\text{-only}=15-5=10\),\quad \(MLE=5\).
Outside \(E\):
\(M\setminus E=25,\ L\setminus E=23\). Since \(|M\cap L|=20\) and it already has \(5\) in \(MLE\) and \(5\) in \(DML\), we get \(ML\text{-only}=10\).
No \(DM\)-only or \(DL\)-only (both would meet \(5\) but all that \(5\) is in \(DML\)). Hence \(\text{Only }M=25-10-5=10\), \(\text{Only }L=23-10-5=8\), \(\text{Only }D=22-5=17\), and \(DML=5\).
At least two sections \(=\) \(ME\text{-only}+LE\text{-only}+ML\text{-only}+MLE+DML\)
\(=25+10+10+5+5=55\).
\[
\boxed{55}
\]